problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 2 Given that the incircle of $\triangle ABC$ touches the sides $BC, CA, AB$ at points $D, E, F$ respectively, and the circumcircle of $\triangle ABC$, $\odot O$, intersects the circumcircles of $\triangle AEF$, $\odot O_{1}$, $\triangle BFD$, $\odot O_{2}$, and $\triangle CDE$, $\odot O_{3}$, at points $A$ and ... | Proof (1) As shown in Figure 3, it is evident that circles $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$ all pass through the incenter of $\triangle ABC$.
(2) As shown in Figure 3, connect $RE$, $RD$, $RA$, and $RB$.
Then $\angle ERD = \angle ECD = \angle ACB = \angle ARB$.
Thus, $\angle ARE = \angle BRD$.
Also, $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,359 |
Example 3 Given a circle $\Gamma$ and a line $l$ that do not intersect, $P, Q, R, S$ are points on the circle $\Gamma$, $PQ$ intersects $RS$, $PS$ intersects $QR$ at points $A, B$ respectively, and $A, B$ lie on the line $l$. Determine all the common points of the circles with $AB$ as their diameter. | Proof As shown in Figure 4, by Corollary 3(1), the circumcircles of $\triangle A S P$ and $\triangle B R S$ intersect at point $K$, and $K$ lies on side $A B$.
Let the center of circle $\Gamma$ be $O$, and its radius be $r$. Then $O K \perp A B$.
By the Round Table Theorem, we have
$$
\begin{array}{l}
B O^{2}-r^{2}=B S... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,360 |
Example 4 Let $A_{1}$ and $C_{1}$ be points on the sides $AB$ and $BC$ of $\square ABCD$, respectively. The line segments $AC_{1}$ and $CA_{1}$ intersect at point $P$. The second intersection point of the circumcircles of $\triangle AA_{1}P$ and $\triangle CC_{1}P$ is $Q$, which lies inside $\triangle ACD$. Prove that ... | Prove that as shown in Figure 5, since the second intersection point of the circumcircles of $\triangle A A_{1} P$ and $\triangle C C_{1} P$ is $Q$, by Corollary 1, $Q$ is the Miquel point of the complete quadrilateral $B C_{1} C P A A_{1}$.
Thus, $A_{1} 、 B 、 C 、 Q$ are concyclic. Therefore, $\angle Q B A=\angle Q B A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,361 |
Example 5 As shown in Figure 6, given that the circumcenter of acute triangle $\triangle ABC$ is $O$, $K$ is a point on side $BC$ (not the midpoint of side $BC$), $D$ is a point on the extension of line segment $AK$, line $BD$ intersects $AC$ at point $N$, and line $CD$ intersects $AB$ at point $M$. Prove: if $OK \perp... | Proof by contradiction.
If points $A, B, D, C$ are not concyclic, let the circumcircle $\odot O$ of $\triangle ABC$ intersect line $AD$ at point $E$, and line $CE$ intersects $AB$ and $BE$ intersects $AC$ at points $P$ and $Q$ respectively.
By Corollary 3(2), the Miquel point $G$ of the complete quadrilateral $PECKAB$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,362 |
Example 6 Convex quadrilateral $ABCD$ is inscribed in $\odot O, BA$ and $CD$ extended meet at point $H$, diagonals $AC$ and $BD$ intersect at point $G$, $O_1$ and $O_2$ are the circumcenters of $\triangle AGD$ and $\triangle BGC$ respectively. Let $O_1O_2$ intersect $OG$ at point $N$, ray $HG$ intersects $\odot O_1$ an... | (2007, IMO China National Training Team Test)
Prove that in Figure 7, draw $GT \perp O_1G$, then $TG$ is tangent to $\odot O$ at point $G$.
Thus, $\angle AGT$
$=\angle ADG$
$=\angle ACB$.
Hence, $TG \parallel BC$.
Therefore,
$O_1G \perp BC$.
And $OO_2 \perp BC$,
Thus, $O_1G \parallel OO_2$.
Similarly,
$OO_1 \parallel ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,363 |
1. Let $A B$ be the diameter of a circle, and draw rays $A D$ and $B E$ on the same side of line $A B$, intersecting at point $C$. If
$$
\angle A E B + \angle A D B = 180^{\circ} \text {, }
$$
then $A C \cdot A D + B C \cdot B E = A B^{2}$. | Let the line $A E$ intersect $B D$ at point $P$. Then $P$, $E$, $C$, and $D$ are concyclic. Following Example 3, we have
\[
\begin{array}{l}
A C \cdot A D + B C \cdot B E \\
= A M \cdot A B + B M \cdot B A = A B^{2} .
\end{array}
\] | A B^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 724,364 |
2. Given that quadrilateral $A B C D$ is inscribed in a circle, the extensions of $A B$ and $D C$ intersect at point $P$, and the extensions of $A D$ and $B C$ intersect at point $Q$. From $Q$, draw two tangents $Q E$ and $Q F$ to the circle, with points of tangency at $E$ and $F$ respectively. Prove that $P$, $E$, and... | Given that the center of the circle is $O$. Then the Miquel point $M$ of the complete quadrilateral $A B P C Q D$ lies on $P Q$, and $O M \perp$ $P Q$. Therefore, points $E, O, F, M, Q$ are concyclic, and $E F$ is the radical axis of $\odot O$ and this circle. Clearly, point $P$ also lies on this radical axis, hence $P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,365 |
3. Given that circle $S_{1}$ intersects with circle $S_{2}$ at points $P$ and $Q$, $A_{1}$ and $B_{1}$ are two points on circle $S_{1}$ different from $P$ and $Q$, lines $A_{1} P$ and $B_{1} P$ intersect circle $S_{2}$ at points $A_{2}$ and $B_{2}$, and line $A_{1} B_{1}$ intersects $A_{2} B_{2}$ at point $C$. Prove: W... | When $A_{1} 、 B_{1}$ vary on circle $S_{1}$, in the complete quadrilateral $C B_{1} A_{1} P B_{2} A_{2}$, the circumcircles of $\triangle A_{1} B_{1} P 、 \triangle A_{2} B_{2} P$ are circles $S_{1}$ and $S_{2}$, which are fixed circles. Their centers $O_{1} 、 O_{2}$ are fixed points, and their Miquel point $Q$ is a fix... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,366 |
4. Given a convex quadrilateral $A B C D, B C=\lambda A D$, and $B C$ is not parallel to $A D$. Let points $E$ and $F$ be on the interiors of sides $B C$ and $A D$, respectively, such that $B E=\lambda D F$. Lines $A C$ and $B D$, $E F$ and $B D$, $E F$ and $A C$ intersect at points $P$, $Q$, and $R$, respectively. Pro... | Suppose the line $A D$ intersects $B C$ at point $S$, and let $O$ be the Miquel point of the complete quadrilateral $S D A P B C$. Then
$$
\begin{array}{l}
\angle O C B=\angle O P B=\angle O A D, \\
\angle B O C=\angle B P C=\angle A P D=\angle A O D .
\end{array}
$$
Thus, $\triangle O C B \backsim \triangle O A D$
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,367 |
Given an integer $n>2$, let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{k} \leqslant 1(k=1,2, \cdots, n)$. Denote
$$
A_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n) .
$$
Prove: $\left|\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} A_{k}\right|<\frac{n-1}{2}$. | The proof process in [1] is relatively loose, so there is room for strengthening the conclusion.
The idea of this paper is to express the supremum of $\left|\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} A_{k}\right|$, and to estimate it using integral inequalities.
$$
\begin{array}{l}
\text { Let }\left|\sum_{k=1}^{n} a_{k}-\su... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,368 |
Example 3 Given 10 points on a circle, color six of them black and the remaining four white. They divide the circumference into arcs that do not contain each other. Rule: arcs with both ends black are labeled with the number 2; arcs with both ends white are labeled with the number $\frac{1}{2}$; arcs with ends of diffe... | Mark all the black points with $\sqrt{2}$, and all the white points with $\frac{1}{\sqrt{2}}$, then the number marked on each arc is exactly the product of the numbers at its two ends. Therefore, the product of the numbers marked on all these arcs is the square of the product of the numbers marked on all the points, i.... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,369 |
Question 1: Find $\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{6}+\cdots\left(n \in \mathbf{N}_{+}\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | By the binomial theorem, we have
$$
(1+x)^{n}=\sum_{k=0}^{n} \mathrm{C}_{n}^{k} x^{k} \text {. }
$$
Substituting $x=1, \omega, \omega^{2}\left(\omega=\mathrm{e}^{\frac{2 \pi x}{3}}\right)$ into the above equation, and adding the resulting three equations on both sides, and noting that
$$
\begin{aligned}
\omega^{3 k}= ... | \frac{1}{3}\left(2^{n}+2 \cos \frac{n \pi}{3}\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,371 |
Question 2: Find the sum of the terms with integer powers of $x$ in the expansion of $(\sqrt[m]{x}+a)^{n}$, where $m, n \in \mathbf{N}, m, n \geqslant 2$, $a \in \mathbf{R}, a \neq 0$.
When $m=a=2$, it is the original problem. | Let the required sum be $S$.
Obviously, when $m>n$, $S=a^{n}$.
When $m \leqslant n$, from the general term of the expansion of $(a+\sqrt[m]{x})^{n}$, $T_{k+1}=\mathrm{C}_{n}^{k} a^{n-k} x^{\frac{k}{m}}$, we know $S=\sum_{m \mid k} \mathrm{C}_{n}^{k} a^{n-k}$.
First, find the sum $S^{\prime}=\sum_{m \mid j} \mathrm{C}_{... | \frac{1}{m} \sum_{k=0}^{m-1}\left(a+\varepsilon^{k}\right)^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,372 |
Question 3 Let $a_{k}$ denote the largest divisor of $k$ that is not divisible by 3. Let $S_{0}=0, S_{k}=a_{1}+a_{2}+\cdots+a_{k}(k \geqslant 1)$, and $A_{n}$ denote the number of integers $k\left(0 \leqslant k<3^{n}\right)$ such that $3 \mid S_{k}$. Prove:
$$
A_{n}=\frac{1}{3}\left(3^{n}+2 \times 3^{\frac{n}{2}} \cos ... | $$
a_{3 k}=a_{k}, a_{3 k+1}=3 k+1, a_{3 k+2}=3 k+2 \text {, }
$$
and $S_{1} \equiv 1(\bmod 3), S_{2} \equiv 0(\bmod 3)$. By mathematical induction, it is easy to prove that for $k \geqslant 0$,
$$
\begin{array}{l}
S_{3 k} \equiv S_{3 k+2} \equiv S_{k}(\bmod 3), \\
S_{3 k+1} \equiv S_{k}+1(\bmod 3) .
\end{array}
$$
Le... | \frac{1}{3}\left(3^{n}+2 \times 3^{\frac{n}{2}} \cos \frac{n \pi}{6}\right) | Number Theory | proof | Yes | Yes | cn_contest | false | 724,373 |
Question 4 Let $P$ be a polynomial of degree $3n$, such that
$$
\begin{array}{l}
P(0)=P(3)=\cdots=P(3 n)=2, \\
P(1)=P(4)=\cdots=P(3 n-2)=1, \\
P(2)=P(5)=\cdots=P(3 n-1)=0 .
\end{array}
$$
$$
\text { If } P(3 n+1)=730 \text {, find } n \text {. }
$$ | Solving, we know
$$
\sum_{k=0}^{3 n+1}(-1)^{k} \dot{\mathrm{C}}_{3 n+1}^{k} P(3 n+1-k)=0,
$$
which means $729+2 \sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{C}_{3 n+1}^{3 j+1}+\sum_{j=0}^{n}(-1)^{3 j} \mathrm{C}_{3 n+1}^{3 j}=0$
Using the multi-section formula, we can find
$$
\begin{array}{l}
\sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,374 |
Question 5 Let $p$ be an odd prime. Find the number of subsets $A$ of the set $\{1,2$, $\cdots, 2 p\}$ that satisfy the following conditions:
(1) $A$ has exactly $p$ elements;
(2) The sum of all elements in $A$ is divisible by $p$. | Let $\sigma(A)$ denote the sum of all elements of $A$, and $[mp]$ denote $\{1,2, \cdots, mp\}$. Then the number of subsets $A$ of $[mp]$ such that $|A|=p$ and $p \mid \sigma(A)$ is
$$
C_{m}=\frac{1}{p}\left[\mathrm{C}_{mp}^{p}+m(p-1)\right].
$$
In fact, consider the polynomial
$$
F(x, y)=(1+x y)\left(1+x y^{2}\right) ... | \frac{1}{p}\left[\mathrm{C}_{2p}^{p}+2(p-1)\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,375 |
1. Find the smallest positive integer $n$, such that there exist $n$ distinct positive integers $s_{1}, s_{2}, \cdots, s_{n}$, satisfying
$$
\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \cdots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010} .
$$ | 1. Suppose the positive integer $n$ satisfies the condition, and let $s_{1}=39$.
Thus, $n \geqslant 39$.
Below, we provide an example to show that $n=39$ satisfies the condition.
Take 39 different positive integers:
$$
2,3, \cdots, 33,35,36, \cdots, 40,67,
$$
which satisfy the given equation.
In conclusion, the minimu... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,376 |
2. Find all pairs of non-negative integers $(m, n)$ such that $m^{2}+2 \times 3^{n}=m\left(2^{n+1}-1\right)$. | 2. For a fixed $n$, the original equation is a quadratic equation in $m$.
When $n=0,1,2$, the discriminant $\Delta \frac{n-2}{3}$.
In particular, $h>1$.
From equation (1), we know
$$
3^{k}\left|\left(2^{n+1}-1\right) \Rightarrow 9\right|\left(2^{n+1}-1\right) \text {. }
$$
Since the order of 2 modulo 9 is 6, it follow... | (m, n)=(6,3),(9,3),(9,5),(54,5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,377 |
3. Find the smallest positive integer $n$, such that there exist rational coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$ | 3. Since $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$, then $n \leqslant 5$.
Thus, it only needs to be proven that $x^{2}+7$ is not equal to the sum of squares of no more than four rational coefficient polynomials.
Assume there exist four rational coefficient polynomials $f_{1}, f_{2}, f_{3}, f_{4}$ (some of which may be 0... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,378 |
4. Let $a, b$ be integers, $P(x) = ax^3 + bx$. For any positive integer $n$, if for all integers $m, k$ we have
$$
n!(P(m) - P(k)) \Rightarrow n!(m - k),
$$
then the pair $(a, b)$ is called “$n$-good”. If the pair $(a, b)$ is $n$-good for infinitely many positive integers $n$, then the pair $(a, b)$ is called “very goo... | 4. (1) First prove: The pair $\left(1,-51^{2}\right)$ is 51-good, but not very good.
Let $p(x)=x^{3}-51^{2} x$.
Since $p(51)=p(0)$, we have
$n \mid(p(51)-p(0))$.
Thus, $n \mid(51-0)$, which means the number of $n$ is finite.
Therefore, $\left(1,-51^{2}\right)$ is not very good.
On the other hand, if $p(m) \equiv p(k) \... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,379 |
Example 4 Remove a $2 \times 2$ square from one corner of an $8 \times 8$ grid paper. Question: Can the remaining 60 squares be covered by 15 pieces of paper shaped like $\square$? | Solve As shown in Figure 1, label the remaining 60 small squares on an $8 \times 8$ grid paper with +1 or -1, then the sum of the numbers in any qualified "four-connected square" is 2 or -2.
Assume these 60 small squares can be divided into 15 qualified four-connected squares, where the sum of the numbers in $x$ of th... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,380 |
1. Find real numbers $a \neq 0, b \neq 0$, such that the graphs of the functions $y=a x^{2}$ and $y=a x+b$ intersect at two distinct points, and the coordinates of the intersection points satisfy the equation
$$
x^{2}-4 a x+y^{2}-2 b y=0 \text {. }
$$ | 1. Let $a, b$ be the desired values.
Then there exist $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right)$ such that
$$
y_{k}=a x_{k}^{2}=a x_{k}+b \quad (k=1,2), \quad x_{1} \neq x_{2},
$$
and for $k=1,2$ we have
$$
\begin{aligned}
0 & =x_{k}^{2}-4 a x_{k}+y_{k}^{2}-2 b y_{k} \\
& =x_{k}^{2}-4 a x_{k}+\left(y_{k}-... | a=2 \pm \sqrt{3}, b=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,382 |
2. As shown in Figure 1, in the acute triangle $\triangle ABC$, $AB > AC$, $\angle BAC = 60^{\circ}$, $O$ and $H$ are the circumcenter and orthocenter of $\triangle ABC$ respectively, and the line $OH$ intersects $AB$ and $AC$ at points $P$ and $Q$ respectively. Prove: $PO = HQ$.
untranslated text:
In the acute trian... | 2. As shown in Figure 3, let the projection of point $O$ on side $AB$ be $N$, and $BH$ intersects $AC$ at point $E$. Then $BE \perp AC$.
In $\triangle EBA$, since $\angle EAB=60^{\circ}$, we have $EA=\frac{AB}{2}=NA$. Also, $\angle NOA=\frac{1}{2} \angle AOB=\angle C=\angle AHE$, thus $\triangle ONA \cong \triangle HE... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,383 |
3. Let $a_{1}, a_{2}, \cdots$ be integers, for any positive integer $n$ we have
$$
a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] \text {. }
$$
If $2001 a_{199}$, find the smallest positive integer $n(n>1)$, such that $200 \mathrm{l} a_{n}$. | 3. From the given, we have
$$
a_{\mathrm{T99}}=198\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right] \text {. }
$$
From $2001 a_{199}$, we get
$$
100 \left\lvert\, 99\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right]\right. \text {. }
$$
Thus, $a_{2}$ is an even number.
Let $a_{2}=2 m$. Then
$$
1001[(m-1)... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,384 |
1. Given $\triangle A B C$ with the three angles $\angle A, \angle B, \angle C$ opposite to the sides of lengths $a, b, c$, and $\angle A=2 \angle B$. Then $\frac{a^{2}}{b(b+c)}=$ $\qquad$ . | 1. 1.
As shown in Figure 4, it is easy to know,
$$
\begin{array}{l}
\triangle A B C \backsim \triangle D A C \\
\Rightarrow \frac{A B}{D A}=\frac{A C}{D C}=\frac{B C}{A C} .
\end{array}
$$
Let $B D=x, D C=y$. Then
$$
\begin{array}{l}
\frac{c}{x}=\frac{b}{y}=\frac{a}{b} \Rightarrow \frac{b+c}{x+y}=\frac{a}{b} \\
\Righ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,385 |
2. Let real numbers $x, y$ satisfy the equation $2 x^{2}+3 y^{2}=6 y$. Then the maximum value of $x+y$ is $\qquad$ . | 2. $1+\frac{\sqrt{10}}{2}$.
Let $x+y=t$. Then $x=t-y$.
From $2(t-y)^{2}+3 y^{2}=6 y$, we get
$5 y^{2}-2(2 t+3) y+2 t^{2}=0$.
Since $y$ is a real number, therefore,
$$
\begin{array}{l}
\Delta=4(2 t+3)^{2}-4 \times 5 \times 2 t^{2} \geqslant 0 \\
\Rightarrow \frac{2-\sqrt{10}}{2} \leqslant t \leqslant \frac{2+\sqrt{10}}... | 1+\frac{\sqrt{10}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,386 |
3. Let $M$ be the midpoint of side $BC$ of $\triangle ABC$, $AB=4$, $AM=1$. Then the minimum value of $\angle BAC$ is $\qquad$ . | 3. $150^{\circ}$.
Let the symmetric point of $A$ with respect to $M$ be $N$. Then quadrilateral $A B N C$ is a parallelogram. Therefore,
$$
\angle B A C=180^{\circ}-\angle A B N \text {. }
$$
Draw a circle with center at $A$ and radius 2. Then when $B N$ is tangent to this circle, $\angle A B N$ is maximized and equa... | 150^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,387 |
4. The solution to the equation $\sqrt{\sqrt{3}-\sqrt{\sqrt{3}+x}}=x$ is | 4. $x=\frac{\sqrt{4 \sqrt{3}-3}-1}{2}$
Let $\sqrt{a+x}=y$, where $a=\sqrt{3}$. Then
$$
\left\{\begin{array} { l }
{ \sqrt { a + x } = y , } \\
{ \sqrt { a - y } = x }
\end{array} \Rightarrow \left\{\begin{array}{l}
a+x=y^{2}, \\
a-y=x^{2} .
\end{array}\right.\right.
$$
Subtracting the two equations, we get
$$
\begin{... | x=\frac{\sqrt{4 \sqrt{3}-3}-1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,388 |
5. The equation $4 x-|3 x-| x+a||=9|x-1|$ has at least one root. Then the range of values for $a$ is $\qquad$ $\therefore$ | 5. $-8 \leqslant a \leqslant 6$.
Transform the original equation into
$$
9|x-1|+|3 x-| x+a||-4 x=0 \text {. }
$$
Consider the continuous function
$$
f(x)=9|x-1|+|3 x-| x+a||-4 x \text {. }
$$
(1) When $x \geqslant 1$, for any expansion of the absolute value, we have
$$
f(x)=9 x-9-4 x \pm 3 x \pm x \pm a=k x+m \text {... | -8 \leqslant a \leqslant 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,389 |
6. As shown in Figure 2, given a convex quadrilateral $ABCD$ satisfying $\angle CAD=45^{\circ}$, $\angle ACD=30^{\circ}$, $\angle BAC$ $=\angle BCA=15^{\circ}$. Then the degree measure of $\angle DBC$ is | $6.90^{\circ}$.
As shown in Figure 5, let $K$ be a point on $CD$ such that $\angle CAK = 30^{\circ}$. Then $BK$ is the perpendicular bisector of $AC$, and
$$
\begin{array}{l}
\angle AKB \\
=\angle BKC \\
=60^{\circ}.
\end{array}
$$
Thus,
$$
\angle AKD = 60^{\circ}.
$$
Since $\angle BAD$
$$
= 60^{\circ}, \text{ theref... | 90^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,390 |
Example 5 Figure 2 is half of a Chinese chess board.
(1) A "knight" jumps $n$ steps back to the starting point, prove: $n$ is even;
(2) Can a knight traverse this half of the chessboard, visiting each square exactly once, and return to the starting point on the last move;
(3) Prove: A knight cannot start from position ... | Solve: Mark the grid points labeled with “ $\times$ ” as +1, and those labeled with “ $O$ ” as -1.
(1) According to the knight's move, its symbol changes once with each step. After $n$ steps, the symbol has changed $n$ times. Since it returns to the starting point, after $n$ changes, the symbol is the same as it was in... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,391 |
7. Given $0<a<b<c<d<500$, and $a+d=b+c$. Also, $bc-ad=93$.
Then the number of ordered quadruples of integers ( $\dot{a}, b, c$,
$d)$ that satisfy the conditions is . $\qquad$ | 7.870.
Since $a+d=b+c$, we set
$$
(a, b, c, d)=(a, a+x, a+y, a+x+y) \text {, }
$$
where $x$ and $y$ are integers, and $0<x<y$.
Then $93=b c-a d$
$$
=(a+x)(a+y)-a(a+x+y)=x y \text {. }
$$
Therefore, $(x, y)=(1,93)$ or $(3,31)$.
First case
$$
(a, b, c, d)=(a, a+1, a+93, a+94) \text {, }
$$
where $a=1,2, \cdots, 405$;... | 870 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,392 |
8. If $a^{2}+b^{2}=1(a, b \neq 0)$,
$$
y=a+b+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b} \text {, }
$$
then the minimum value of $|y|$ is | 8. $2 \sqrt{2}-1$.
Notice that $y=a+b+\frac{1}{a b}+\frac{a+b}{a b}$.
Let $t=a+b$. Then, $a b=\frac{1}{2}\left(t^{2}-1\right)$, where $|t| \leqslant \sqrt{2}$, and $t \neq 1$.
Thus, $y=t+\frac{2}{t^{2}-1}+\frac{2 t}{t^{2}-1}$
$$
=t+\frac{2}{t-1}=t-1+\frac{2}{t-1}+1 \text {. }
$$
When $t>1$, $t-1+\frac{2}{t-1} \geqsla... | 2 \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,393 |
9. The number of prime pairs $(a, b)$ that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$
is $\qquad$. | 9.2.
If $a=b$, then the equation is equivalent to $a^{2a}=(3a+1)^2$. But $a \mid a^{2a}, a \mid (3a+1)^2$, which is a contradiction, so $a \neq b$.
Assume without loss of generality that $a > b$. Since the difference between the two factors on the right side of the original equation is $a-b$, $a$ can only divide one o... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,394 |
10. Xiao Li and Xiao Zhang are running on a circular track at a uniform speed. They start at the same time and place. Xiao Li runs clockwise, completing a lap every 72 seconds; Xiao Zhang runs counterclockwise, completing a lap every 80 seconds. If a $\frac{1}{4}$ circular arc interval is marked on the track with the s... | $10.3,9,11,18$.
Let the start time be 0. Then the time intervals during which Xiao Li and Xiao Zhang run in the designated area are
$$
[0,9] \text{ and } [72k-9,72k+9]
$$
and $[0,10] \text{ and } [80k-10,80k+10]$,
where $k=1,2, \cdots$.
Obviously, $[0,9]$ is the common part of the two types of time intervals in the fi... | 3,9,11,18 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,395 |
1. Given $\min _{x \in R} \frac{a x^{2}+b}{\sqrt{x^{2}+1}}=3$.
(1) Find the range of $b$;
(2) For a given $b$, find $a$. | 1. Solution 1 (1) Let $f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}}$.
Then $f(0)=b \geqslant 3$. Furthermore, $a>0$.
(i) When $b-2 a \geqslant 0$,
$$
\begin{array}{l}
f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}} \\
=a \sqrt{x^{2}+1}+\frac{b-a}{\sqrt{x^{2}+1}} \\
\geqslant 2 \sqrt{a(b-a)}=3 .
\end{array}
$$
The equality holds if and... | a=\frac{b-\sqrt{b^{2}-9}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,396 |
2. Given that $a$, $b$, and $c$ are pairwise coprime positive integers, and satisfy
$$
a^{2} \mid \left(b^{3} + c^{3}\right), \quad b^{2} \mid \left(a^{3} + c^{3}\right), \quad c^{2} \mid \left(a^{3} + b^{3}\right) \text{. }
$$
Find the values of $a$, $b$, and $c$.
(Supplied by Yang Xiaoming) | 2. From the given conditions, we have
$$
\begin{array}{l}
a^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), b^{2} \mid \left(a^{3}+b^{3}+c^{3}\right), \\
c^{2} \mid \left(a^{3}+b^{3}+c^{3}\right).
\end{array}
$$
Since \(a\), \(b\), and \(c\) are pairwise coprime, we have
$$
a^{2} b^{2} c^{2} \mid \left(a^{3}+b^{3}+c^{3}\righ... | (1,1,1),(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,2,1),(3,1,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,397 |
3. Find all positive integers $n$ such that any 35-element subset of the set
$$
M=\{1,2, \cdots, 50\}
$$
contains at least two distinct elements $a, b$ satisfying $a+b=n$ or $a-b=n$. | 3. Let $A=\{1,2, \cdots, 35\}$.
Then for any $a, b \in A$, we have
$$
a-b, a+b \leqslant 34+35=69.
$$
On the other hand, let $A=\left\{a_{1}, a_{2}, \cdots, a_{35}\right\}$, and
$$
a_{1}<a_{2}<\cdots<a_{35}.
$$
(1) When $1 \leqslant n \leqslant 19$, consider
$$
\begin{array}{l}
1 \leqslant a_{1}<a_{2}<\cdots<a_{35} \... | 1 \text{ to } 69 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,398 |
4. Draw any line through the circumcenter $O$ of $\triangle ABC$, intersecting sides $AB$ and $AC$ at points $M$ and $N$ respectively. Let $E$ and $F$ be the midpoints of $BN$ and $CM$ respectively. Prove that $\angle EOF = \angle A$. | 4. First, prove that the conclusion holds for any triangle.
Consider three cases.
For the right triangle $\triangle ABC$, the conclusion is obvious.
Indeed, as shown in Figure 2, if $\angle ABC$ is a right angle, then the circumcenter $O$ is the midpoint of the hypotenuse $AC$.
A line through $O$ intersects $AB$ and $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,399 |
5. As shown in Figure 1, let $A A_{0}$, $B B_{0}$, and $C C_{0}$ be the angle bisectors of $\triangle A B C$. Draw $A_{0} A_{1} / / B B_{0}$ and $A_{0} A_{2} / / C C_{0}$, where points $A_{1}$ and $A_{2}$ lie on $A C$ and $A B$ respectively. The line $A_{1} A_{2}$ intersects $B C$ at point $A_{3}$; similarly, obtain po... | 5. By the converse of Menelaus' theorem, it suffices to prove:
$$
\frac{A B_{3}}{B_{3} C} \cdot \frac{C A_{3}}{A_{3} B} \cdot \frac{B C_{3}}{C_{3} A}=1 \text {. }
$$
From the line $A_{1} A_{2} A_{3}$ intersecting $\triangle A B C$, we get
$$
\begin{array}{l}
\frac{C A_{3}}{A_{3} B} \cdot \frac{B A_{2}}{A_{2} A} \cdot ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,400 |
6. Let $P_{1}, P_{2}, \cdots, P_{n}$ be $n$ fixed points on a plane, and $M$ be any point on the line segment $A B$ in the plane. Denote $\left|P_{i} M\right|$ as the distance between point $P_{i}(i=1,2, \cdots, n)$ and $M$. Prove:
$$
\sum_{i=1}^{n}\left|P_{i} M\right| \leqslant \max \left\{\sum_{i=1}^{n}\left|P_{i} A\... | 6. Let the origin be $O$. Then
$$
\begin{array}{l}
\overrightarrow{O M}=t \overrightarrow{O A}+(1-t) \overrightarrow{O B}(t \in(0,1)), \\
\left|P_{i} M\right|=\left|\overrightarrow{O M}-\overrightarrow{O P_{i}}\right| \\
=\left|t \overrightarrow{O A}+(1-t) \overrightarrow{O B}-t \overrightarrow{O P_{i}}-(1-t) \overrigh... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,401 |
Example 6 Figure 3 (a) is
an electronic display of English letters, where each operation can simultaneously change 4 letters in a row or 4 letters in a column. The rule for change is: according to the order of the English alphabet, each letter changes to the next one (i.e., $A$ changes to $B, B$ changes to $C, \cdots ... | For the 26 English letters, mark them with numbers 1 and -1 alternately according to the alphabetical order. Thus, Figure 3 (a) and (b) become Figure 4 (a) and (b), respectively.
According to the problem, one operation on Figure 3 (a) is equivalent to performing the following operation on Figure 4 (a): multiplying all... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,402 |
7. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=a_{2}=1, a_{n}=7 a_{n-1}-a_{n-2}(n \geqslant 3) \text {. }
$$
Prove: For each $n \in \mathbf{N}_{+}, a_{n}+a_{n+1}+2$ is a perfect square.
(Tao Pingsheng) | 7. The first few terms of the sequence can be easily found to be
$$
1,1,6,41,281,1926, \cdots \text {. }
$$
Notice that
$$
\begin{array}{r}
a_{1}+a_{2}+2=2^{2}, a_{2}+a_{3}+2=3^{2}, \\
a_{3}+a_{4}+2=7^{2}, a_{4}+a_{5}+2=18^{2},
\end{array}
$$
Construct the sequence $\left\{x_{n}\right\}$ :
$$
\begin{array}{l}
x_{1}=2... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,403 |
1. Prove: There exist two functions $f, g: \mathbf{R} \rightarrow \mathbf{R}$, such that the function $f(g(x))$ is strictly decreasing on $\mathbf{R}$, while $g(f(x))$ is strictly increasing on $\mathbf{R}$. | 1. Let $A=\underset{k \in \mathrm{Z}}{ }\left(\left[-2^{2 k+1},-2^{2 k}\right) \cup\left(2^{2 k}, 2^{2 k+1}\right]\right)$, $B=\bigcup_{k \in \mathbb{Z}}\left(\left[-2^{2 k},-2^{2 k-1}\right) \cup\left(2^{2 k-1}, 2^{2 k}\right]\right)$. Then $A=2 B, B=2 A, A=-A, B=-B, A \cap B=\varnothing$, and $A \cup B \cup\{0\}=\mat... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,405 |
3. Let circle $\omega$ be the circumcircle of $\triangle A B C$, a moving line $l$ parallel to $B C$ intersects segments $A B$ and $A C$ at points $D$ and $E$, respectively, and intersects circle $\omega$ at points $K$ and $L$ (point $D$ lies between $K$ and $E$), $\Gamma_{1}$ is the circle tangent to segments $K D$, $... | 3. Let $P$ be the intersection of the internal common tangents of circles $\Gamma_{1}$ and $\Gamma_{2}$, and let line $m$ be the angle bisector of $\angle BAC$. Since $KL \parallel BC$, $m$ is also the angle bisector of $\angle KAL$.
First, perform a reflection about line $m$, then perform an inversion with center $A$... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,407 |
4. For a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, let $\Omega(n)=\sum_{i=1}^{s} \alpha_{i}$ be the number of all prime factors of $n$, where the prime factors are counted with multiplicity. Define $\lambda(n)=(-1)^{\Omega(n)}$ (for example, $\lambda(12)=$ $\left.\lambda\left(2^{2} \times 3\right)=(-1)^{... | 4. Notice that, for any positive integers $m, n$, we have
$$
\Omega(m n)=\Omega(m)+\Omega(n),
$$
which means $\Omega$ is a completely additive function.
Therefore, $\lambda(m n)=\lambda(m) \lambda(n)$, which means $\lambda$ is a completely multiplicative function.
Hence, for any prime $p$ and positive integer $k$, we ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,408 |
5. For each positive integer $n(n \geqslant 3)$, determine the relationship between $n$ distinct points $X_{1}, X_{2}, \cdots, X_{n}$ in the plane that satisfy the following property:
For any pair of distinct points $X_{i}, X_{j}$, there exists a permutation $\sigma$ of $\{1,2, \cdots, n\}$ such that for all $k$ $(1 \... | 5. First, establish an appropriate Cartesian coordinate system so that the vector $x_{k}$ from the origin to point $X_{k}$ satisfies
$$
\begin{array}{l}
\frac{1}{n} \sum_{k=1}^{n} x_{k}=0 . \\
\text { By } d\left(X_{i}, X_{k}\right)^{2}=\left\|x_{i}-x_{k}\right\|^{2} \\
=\left(x_{i}-x_{k}\right) \cdot\left(x_{i}-x_{k}\... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,409 |
6. Each cell of a $2011 \times 2011$ grid is labeled with an integer from $1,2, \cdots, 2011^{2}$, such that each number is used exactly once. Now, the left and right boundaries, as well as the top and bottom boundaries of the grid, are considered the same, forming a torus (which can be viewed as the surface of a "doug... | 6. Let $N=2011$.
Consider a general $N \times N$ table.
When $N=2$, the conclusion is obvious, and the required $M=2$. An example is shown in Table 1.
Table 1
\begin{tabular}{|l|l|}
\hline 1 & 2 \\
\hline 3 & 4 \\
\hline
\end{tabular}
When $N \geqslant 3$, first prove:
$M \geqslant 2 N-1$.
Starting from a state where... | 4021 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,410 |
10.1. A table of numbers consisting of $n$ rows and 10 columns, with each element being an integer from $0 \sim 9$, satisfies the following condition: for any row $A$ and any two columns $B$ and $C$, there exists another row $D$ such that $D$ differs from $A$ only in the numbers in columns $B$ and $C$. Prove: $n \geqsl... | 10.1. Let $R_{0}$ be the first row. Arbitrarily select $2 m$ columns $C_{1}$, $C_{2}, \cdots, C_{2 m}$. By the given condition, there exists a row $R_{1}$, which differs from $R_{0}$ only in $C_{1}$ and $C_{2}$; further, there exists a row $R_{2}$, which differs from $R_{1}$ only in $C_{3}$ and $C_{4}$; $\cdots \cdots$... | 512 | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,411 |
10.2. Nine real-coefficient quadratic polynomials
$$
x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \cdots, x^{2}+a_{9} x+b_{9}
$$
satisfy: The sequences $a_{1}, a_{2}, \cdots, a_{9}$ and $b_{1}, b_{2}, \cdots, b_{9}$ are both arithmetic sequences. It is known that the polynomial obtained by adding these nine polynomials h... | $$
\begin{array}{l}
\text { 10. 2. Let } P_{i}(x)=x^{2}+a_{i} x+b_{i}(i=1,2, \cdots, 9), \\
P(x)=P_{1}(x)+P_{2}(x)+\cdots+P_{9}(x) .
\end{array}
$$
Notice that
$$
P_{i}(x)+P_{10-i}(x)=2 P_{5}(x)=\frac{2}{9} P(x)
$$
has real roots, denoted as $x_{0}$.
Then $P_{i}\left(x_{0}\right)+P_{10-i}\left(x_{0}\right)=0$.
Thus, ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,412 |
Example 7 There are $p(p>100)$ cups on the table, all with their mouths facing up. Operate on the cups according to the following rules: the 1st time, arbitrarily flip 1 cup, the 2nd time, arbitrarily flip 2 cups, $\cdots \cdots$ the $n$th time, arbitrarily flip $n(n \leqslant p)$ cups. Each operation changes the direc... | Prove that assigning a value of +1 to a cup with its opening facing up and -1 to a cup with its opening facing down, the product of the values of all cups before any operation is $a_{0}=1$. Let the product of the values of all cups after the $n$-th operation be $a_{n}$.
Since flipping a cup is equivalent to multiplyin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,413 |
10.3. If after dividing a group of people into $k$ groups arbitrarily, there is at least one group in which two people know each other, then this group of people is called “$k$-indivisible”. It is known that in a group of “3-indivisible” people, there do not exist four people who all know each other. Prove: This group ... | 10.3. First, convert the problem into a graph theory problem.
Each person corresponds to a vertex in the graph. If two people know each other, then the corresponding vertices are connected by an edge. Thus, we obtain the graph \( G \), and graph \( G \) cannot be 3-colored such that each edge connects two vertices of ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,414 |
10.4. Given that the perimeter of $\triangle A B C$ is 4, take points $X, Y$ on the rays $A B, A C$ respectively, such that $A X=A Y=1$. The segment $B C$ intersects $X Y$ at point $M$. Prove: one of the triangles $\triangle A B M$ and $\triangle A C M$ has a perimeter of 2. | 10.4. Since line segments $B C$ and $X Y$ intersect, without loss of generality, assume
$$
A B>A X, A C<A Y \text {. }
$$
Let the excircle of $\triangle A B C$, denoted by $\Gamma$, touch side $B C$ at point $R$, and touch the extensions of $A B$ and $A C$ at points $P$ and $Q$, respectively. Then
$$
A P=A Q=\frac{1}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,415 |
10.5. Given 10 distinct real numbers. For any two numbers $a$ and $b$, Watt writes the number $(a-b)^{2}$ in his notebook, and Betya writes the number $\left|a^{2}-b^{2}\right|$ in her notebook. Question: Can the 45-element unordered arrays obtained from the 45 numbers written by each of them be equal? | 10.5. Not possible.
Assume by contradiction that there exist such 10 numbers. If there is a 0 among them, removing the 0 still leaves a 36-element array from the remaining 9 numbers, and the conclusion still holds.
Now assume there are 9 or 10 distinct non-zero numbers, and let \(a\) and \(b\) represent the smallest ... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,416 |
10.6. Given that $B B_{1}$ and $C C_{1}$ are the altitudes of the acute $\triangle A B C$, and points $P$ and $Q$ are taken on the extensions of $B B_{1}$ and $C C_{1}$ respectively such that $\angle P A Q=90^{\circ}$, and $A F$ is the altitude of $\triangle A P Q$. Prove: $\angle B F C=90^{\circ}$. | 10.6. Points $B_{1}$ and $C_{1}$ lie on the circle with diameter $BC$. We only need to prove that point $F$ also lies on this circle, i.e., to prove that points $C$, $B_{1}$, $F$, and $C_{1}$ are concyclic.
In fact, since $\angle A B_{1} P = \angle A F P = 90^{\circ}$, points $B_{1}$ and $F$ lie on the circle with dia... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,417 |
10.7. For any positive integers $a, b (a>b>1)$, define the sequence $x_{n}=\frac{a^{n}-1}{b^{n}-1}(n=1,2, \cdots)$. It is known that the defined sequence does not have $d$ consecutive terms consisting of prime numbers. Find the minimum value of $d$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The la... | 10.7. The minimum value of $d$ is 3.
When $a=4, b=2$, $x_{1}=3, x_{2}=5$ are prime numbers.
Therefore, the minimum value of $d$ is greater than 2.
Next, we prove: there cannot be three consecutive terms all being prime numbers.
In fact, a stronger conclusion can be proven:
For $n \geqslant 2, x_{n}$ and $x_{n+1}$ cann... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,418 |
10.8. A $2010 \times 2010$ grid is divided into some L-shaped pieces (an L-shaped piece is obtained by removing one $1 \times 1$ square from a $2 \times 2$ square). Prove: It is possible to select one square from each L-shaped piece so that the number of selected squares in each row and each column of the grid is the s... | 10.8. Let $n=\frac{2010}{3}=670$. Number the rows from top to bottom and the columns from left to right. We need to prove: it is possible to select one square in each triangular shape such that for $k=1,2, \cdots, 2010$, the first $k$ rows and the first $k$ columns of the table contain exactly $k n$ selected squares. T... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,419 |
1. Given the equation about $x$
$$
|| x-2|-| x+7||=k
$$
has two real roots. Then the range of $k$ is ( ).
(A) $k \geqslant 0$
(B) $0 \leqslant k<9$
(C) $0<k<9$
(D) $0<k \leqslant 9$ | - 1. C.
From the problem, we know $k \geqslant 0$. Then
$|x-2|-|x+7|= \pm k$.
Let $|x-2|-|x+7|=p$. Then
$$
p=\left\{\begin{array}{ll}
-9, & x \geqslant 2 ; \\
9, & x \leqslant-7 ; \\
-2 x-5, & -7<x<2 .
\end{array}\right.
$$
Therefore, when $-9<p<9$,
$$
|x-2|-|x+7|=p
$$
has a unique solution.
Thus, $-9<k<9,-9<-k<9$.
... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,420 |
2. Let $a, b, c$ be non-zero real numbers that are not all equal. If
$$
\begin{array}{l}
x=(a+b+c)^{2}-9 a b, \\
y=(a+b+c)^{2}-9 b c, \\
z=(a+b+c)^{2}-9 c a,
\end{array}
$$
then $x, y \backslash z$ ( ).
(A) are all not less than 0
(B) are all not greater than 0
(C) at least one is less than 0
(D) at least one is great... | 2. D.
Notice that
$$
\begin{array}{l}
x+y+z=3(a+b+c)^{2}-9(a b+b c+c a) \\
=3\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\
=\frac{3}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]>0 .
\end{array}
$$
Therefore, at least one of $x, y, z$ is greater than 0. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,421 |
3. Given a right-angled triangle with the two legs $a, b$ $(a \leqslant b)$ and the hypotenuse $c$ all being integers, and the radius of its inscribed circle $r=3$. Then the number of such right-angled triangles is ( ).
(A) 0
(B) 1
(C) 3
(D) infinitely many | 3. C.
From $a+b-c=2 r=6$, we get
$$
a^{2}+b^{2}=c^{2}=(a+b-6)^{2} \text {. }
$$
Rearranging gives $(a-6)(b-6)=18$.
It is clear that $a$ and $b$ are both positive integers greater than 6. Thus,
$$
(a-6, b-6)=(1,18),(2,9),(3,6) \text {. }
$$
Solving gives
$$
(a, b, c)=(7,24,25),(8,15,17),(9,12,15) \text {. }
$$
There... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,422 |
4. Given the quadratic equation $x^{2}+p x+q=0$ with roots $x_{1} 、 x_{2}$. The conditions are:
(1) $p 、 q 、 x_{1} 、 x_{2}$ are all integers that leave a remainder of 1 when divided by 3;
(2) $p 、 q 、 x_{1} 、 x_{2}$ are all integers that leave a remainder of 2 when divided by 3;
(3) $p 、 q$ are integers that leave a re... | 4. A.
From $p+x_{1}+x_{2}=0$ we know that the equation does not satisfy conditions (3) and (4). From $q=x_{1} x_{2}$ we know that the equation does not satisfy condition (2). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,423 |
Example $8 A, B, C, D, E$ five people participated in an exam, which has seven questions, all of which are true or false questions. The scoring criteria are: for each question, 1 point is awarded for a correct answer, 1 point is deducted for a wrong answer, and no points are awarded or deducted for unanswered questions... | Let $x_{k}=\left\{\begin{array}{ll}1, & \text { if the conclusion of the } k \text {th question is correct; } \\ -1, & \text { if the conclusion of the } k \text {th question is incorrect, }\end{array}\right.$ where $k=1,2, \cdots, 7$.
At this point, if the conclusion is judged to be correct (i.e., marked with a “ $\c... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,424 |
5. In a sports meet, ten athletes participate in a table tennis competition, where each pair of athletes plays exactly one match. During the competition, the first athlete wins $x_{1}$ games and loses $y_{1}$ games; the second athlete wins $x_{2}$ games and loses $y_{2}$ games; $\cdots \cdots$ the tenth athlete wins $x... | 5. C.
Given that $x_{i}+y_{i}=9(i=1,2, \cdots, 10)$, and $x_{1}+x_{2}+\cdots+x_{10}=y_{1}+y_{2}+\cdots+y_{10}=45$.
Then $M-N$
$$
\begin{aligned}
= & \left(x_{1}^{2}-y_{1}^{2}\right)+\left(x_{2}^{2}-y_{2}^{2}\right)+\cdots+\left(x_{10}^{2}-y_{10}^{2}\right) \\
= & \left(x_{1}-y_{1}\right)\left(x_{1}+y_{1}\right)+\left(... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,425 |
6. As shown in Figure 1, in the right triangle $\triangle ABC$, it is known that $O$ is the midpoint of the hypotenuse $AB$, $CD \perp AB$ at point $D$, and $DE \perp OC$ at point $E$. If the lengths of $AD$, $DB$, and $CD$ are all rational numbers, then among the line segments $OD$, $OE$, $DE$, and $AC$, the length th... | 6. D.
Since the lengths of $A D$, $D B$, and $C D$ are all rational numbers, we have
$$
O A=O B=O C=\frac{A D+B D}{2}
$$
which is a rational number.
Thus, $O D=O A-A D$ is a rational number.
Therefore, $O E=\frac{O D^{2}}{O C}$ and $D E=\frac{D C \cdot D O}{O C}$ are both rational numbers, while $A C=\sqrt{A D \cdot ... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,426 |
2. Let real numbers $a, b, c, d, e$ satisfy $a<b<c<d<e$, and among the 10 sums of any two numbers, the smallest three sums are $32, 36, 37$, and the largest two sums are $48, 51$. Then
$$
e=
$$
$\qquad$ | 2.27.5.
From the problem, we know
$$
\begin{array}{l}
a+b=32, a+c=36, \\
c+e=48, d+e=51 .
\end{array}
$$
It is easy to see that, $c-b=4, d-c=3, d-b=7$.
Thus, $a+d=(a+b)+(d-b)=39$.
Therefore, $b+c=37$.
Then $2 a=(a+b)+(a+c)-(b+c)=31$.
So $a=15.5, b=16.5, c=20.5$,
$$
d=23.5, e=27.5 \text {. }
$$
Upon verification, the... | 27.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,428 |
3. If the length, width, and height of a rectangular prism are all prime numbers, and the sum of the areas of two adjacent sides is 341, then the volume of this rectangular prism $V=$ $\qquad$ . | 3. 638.
Let the length, width, and height of the rectangular prism be $x, y, z$. From the problem, we have
$$
\begin{array}{l}
x(y+z)=341=11 \times 31 \\
\Rightarrow(x, y+z)=(11,31),(31,11) .
\end{array}
$$
Since $y+z$ is odd, one of $y, z$ must be 2 (let's assume $z=2$).
Also, $11-2=9$ is not a prime number, so
$$
(... | 638 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,429 |
4. Given ten distinct rational numbers, among them, the sum of any nine is an irreducible proper fraction with a denominator of 22 (a proper fraction where the numerator and denominator have no common divisor). Then the sum of these ten rational numbers $S=$ $\qquad$ | 4. $\frac{5}{9}$.
The irreducible proper fraction with a denominator of 22 is $\frac{i}{22}$, where $i=1,3,5,7,9,13,15,17,19,21$.
According to the problem, ten distinct rational numbers $a_{1}, a_{2}, \cdots, a_{10}$, each time taking nine of them, the sum $S-a_{i}$ $(i=1,2, \cdots, 10)$ should be one of the ten irre... | \frac{5}{9} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,430 |
One, (20 points) Given real numbers $a, b, c, d$ satisfy $a+b+c+d=4$, $ab+ac+ad+bc+bd+cd=-\frac{14}{3}$. Find the maximum and minimum values of $b+c+d$.
保留源文本的换行和格式，直接输出翻译结果。 | Let's assume $b+c+d=t$. Then
$$
\begin{array}{l}
a=4-t, \\
b c+b d+c d=-\frac{14}{3}-a t \\
=-\frac{14}{3}-(4-t) t=t^{2}-4 t-\frac{14}{3} .
\end{array}
$$
Notice that
$$
\begin{array}{l}
2(b+c+d)^{2}-6(b c+b d+c d) \\
=(b-c)^{2}+(b-d)^{2}+(c-d)^{2} \geqslant 0,
\end{array}
$$
i.e., $2 t^{2}-6\left(t^{2}-4 t-\frac{14}... | 7 \text{ and } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,431 |
II. (25 points) As shown in Figure 3, given that $P A$ and $P B$ are tangent to $\odot O$ at points $A$ and $B$, respectively, and a secant line through point $P$ intersects $\odot O$ at points $C$ and $D$. Line $C E$ parallel to $P A$ intersects chord $A B$ at point $E$, and the extension of $D E$ intersects $P A$ at ... | II. As shown in Figure 5, connect $O P, O B, E Q, B Q, B C$, and $A D$, and extend $C E$ to intersect $A D$ at point $F$. Draw $O Q \perp C D$ at point $Q$. Then, $C Q = Q D$.
Since $P A$ and $P B$ are tangents to $\odot O$, we have
$O B \perp P B, \angle P A B = \angle P O B$.
Also, $\angle O Q P = \angle O B P = 90^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,432 |
Three. (25 points) Find the real number $k(k \neq \pm 1)$, such that the equation
$$
\left(k^{2}-1\right) x^{2}-6(3 k-1) x+72=0
$$
has two positive integer roots. | $$
x_{1}=\frac{12}{k+1}, x_{2}=\frac{6}{k-1} \text {. }
$$
Given that $x_{1}, x_{2}$ are positive integers, $k$ must be a rational number greater than 1. Let $k=\frac{m}{n}(m$ and $n$ are coprime, $m>n>0)$. Then
$$
\begin{array}{l}
x_{1}=\frac{12 n}{m+n}, x_{2}=\frac{6 n}{m-n} . \\
\text{Since } (m+n, n)=(m-n, n)=1, \... | k=2,3, \frac{7}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,433 |
In $\triangle A B C$, points $D, E, F$ are on sides $B C, C A, A B$ respectively, and let $M$ be their Miquel point. When $A D \perp B C$, and $M$ lies on line $A D$, the points $E, F$ and the circumcenters $O_{1}, O_{2}$ of $\triangle B D F$ and $\triangle D C E$ are concyclic if and only if $M$ is the orthocenter of ... | From $\angle M E C=180^{\circ}-\angle M D C=90^{\circ}$, we know
$M E \perp A C$.
Similarly, $M F \perp A B$.
From $A F \cdot A B=A M \cdot A D=A E \cdot A C$, we know $B, C, E, F$ are concyclic.
Since $A D \perp B C$, then $O_{1}, O_{2}$ are the midpoints of $B M, C M$ respectively, which means $\mathrm{O}_{1} \mathr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,435 |
2. Let
$$
\begin{array}{l}
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \\
\text { then } f(1)+f(2)+\cdots+f(50)=
\end{array}
$$ | $$
\begin{array}{l}
x^{2}-53 x+196=(x-4)(x-49) . \\
\text { Therefore, when } 4 \leqslant x \leqslant 49, f(x)=0 . \\
\text { Then } f(1)+f(2)+\cdots+f(50) \\
=f(1)+f(2)+f(3)+f(50)=660 \text {. }
\end{array}
$$ | 660 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,436 |
3. Let $t=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}$. Then the sum of all real solutions of the equation $(t-1)(t-2)(t-3)=0$ with respect to $x$ is $\qquad$ | 3.4.
Notice that
$$
f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}
$$
is a monotonically decreasing function.
When $x=0,1,3$, its values are $3, 2, 1$ respectively, so the three roots of the equation are $x=0,1,3$, and their sum is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,437 |
4. The highest point of the ellipse obtained by rotating the ellipse $\frac{x^{2}}{2}+y^{2}=1$ counterclockwise by $45^{\circ}$ around the coordinate origin is $\qquad$ .
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. $\frac{\sqrt{15}}{3}$.
The tangent line to the ellipse at its highest point has a slope of 0. After rotating the ellipse clockwise by $45^{\circ}$ to restore it, the slope of its tangent line is -1.
The point of tangency can be found to be $\left(\frac{2 \sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)$.
Therefore, the requ... | \frac{\sqrt{15}}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 724,438 |
5. Two boxes are filled with black and white balls. The total number of balls in both boxes is 25. Each time, a ball is randomly drawn from each box. The probability that both balls are black is $\frac{27}{50}$, and the probability that both balls are white is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$.... | 5.26.
Let the first box contain $x$ balls, of which $p$ are black, and the second box contain $25-x$ balls, of which $q$ are black. Then
$$
\begin{array}{l}
\frac{p}{x} \cdot \frac{q}{25-x}=\frac{27}{50}, \\
50 p q=27 x(25-x) .
\end{array}
$$
Thus, $x$ is a multiple of 5.
Substituting $x=5$ and $x=10$ into the two e... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,439 |
6. Let $0 \leqslant x \leqslant 8$. Then
$$
f(x)=\frac{\sqrt{x\left(x^{2}+8\right)(8-x)}}{x+1}
$$
the range of values is | 6. $[0,4]$.
Notice that
$$
\begin{array}{l}
\frac{\sqrt{x\left(x^{2}+8\right)(8-x)}}{x+1} \\
=\frac{\sqrt{\left(x^{2}+8\right)\left(8 x-x^{2}\right)}}{x+1} \\
\leqslant \frac{\left(x^{2}+8\right)+\left(8 x-x^{2}\right)}{2(x+1)}=4 .
\end{array}
$$
When $x=2$, the equality holds. | [0,4] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,440 |
7. For a regular quadrilateral pyramid $P-ABCD$ with base and lateral edge lengths all being $a$, $M$ and $N$ are moving points on the base edges $CD$ and $CB$ respectively, and $CM = CN$. When the volume of the tetrahedron $P-AMN$ is maximized, the angle between line $PA$ and plane $PMN$ is $\qquad$ | 7. $\frac{\pi}{4}$.
Let $C M=x(0 \leqslant x \leqslant a)$. Then
$$
\begin{array}{l}
S_{\triangle A M N}=a x-\frac{1}{2} x^{2} \\
=\frac{1}{2} x(2 a-x) \leqslant \frac{1}{2} a^{2} .
\end{array}
$$
When $x=a$, that is, when point $M$ coincides with $D$, the equality holds above.
Given $A C \perp$ plane $P B D$, the a... | \frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,441 |
8. Given that the three sides of $\triangle A B C$ are $13, 14$, and $15$. Four circles $\odot \mathrm{O}, \odot \mathrm{O}_{1}, \odot \mathrm{O}_{2}$, and $\odot O_{3}$ with the same radius $r$ are placed inside $\triangle A B C$, and $\odot O_{1}$ is tangent to sides $A B$ and $A C$, $\odot O_{2}$ is tangent to sides... | 8. $\frac{260}{129}$.
Let's assume $a=13, b=14, c=15$.
It is easy to know that $\triangle A B C \sim \triangle \mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$, and $O$ is the circumcenter of $\triangle O_{1} O_{2} O_{3}$, with the circumradius being $2 r$.
Then $\cos \angle O_{1} O_{3} O_{2}=\cos C=\frac{5}{13}$,
$$
\si... | \frac{260}{129} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,442 |
9. (16 points) Given
$$
f(x)=\sin 2x + 3 \sin x + 3 \cos x \quad (0 \leqslant x < 2 \pi) \text{.}
$$
(1) Find the range of $f(x)$;
(2) Find the intervals of monotonicity of $f(x)$. | (1) Let
$\sin x+\cos x=t(-\sqrt{2} \leqslant t \leqslant \sqrt{2})$.
Then $\sin 2 x=t^{2}-1$.
Thus, we only need to find the range of $g(t)=t^{2}+3 t-1$.
Also, $g(t)=\left(t+\frac{3}{2}\right)^{2}-\frac{13}{4}$, so when $t= \pm \sqrt{2}$,
$g(t)$ attains its extreme values, which means the range of $f(x)$ is
$$
[1-3 \sq... | \left[0, \frac{\pi}{4}\right],\left[\frac{5 \pi}{4}, 2 \pi\right] \text{ for increasing; } \left[\frac{\pi}{4}, \frac{5 \pi}{4}\right] \text{ for decreasing} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,443 |
10. (20 points) Given the sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}=\frac{1}{3}, x_{n+1}=x_{n}^{2}+x_{n}(n=1,2, \cdots) \text {. }
$$
Find the value of $\sum_{n=1}^{2002} \frac{1}{x_{n}+1}$ between which two integers. | 10. Notice that $x_{n}^{2}+x_{n}=x_{n}\left(x_{n}+1\right)$.
$$
\begin{array}{l}
\text { Then } \frac{1}{x_{n}\left(x_{n}+1\right)}=\frac{1}{x_{n}}-\frac{1}{x_{n}+1} \\
\Rightarrow \frac{1}{x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n}+1} \\
\Rightarrow \frac{1}{x_{n}+1}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\text { Hence }... | 2,3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,444 |
11. (20 points) Through the point $(2,3)$, draw a moving line $l$ intersecting the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at two distinct points $P$ and $Q$. Draw the tangents to the ellipse at $P$ and $Q$, and let the intersection of these tangents be $M$.
(1) Find the equation of the locus of point $M$;
(2) Let $O$ be the... | 11. (1) According to the problem, let the equation of line $l$ be
$$
y=k(x-2)+3 \text {, }
$$
Combining with the ellipse equation, we get
$$
\begin{array}{l}
\left(1+4 k^{2}\right) x^{2}+8 k(3-2 k) x+4\left(4 k^{2}-12 k+8\right) \\
=0 .
\end{array}
$$
From $\Delta=64(3 k-2)>0$, we get $k>\frac{2}{3}$.
Let $P\left(x_{... | x-y+1=0 \text { or } 11 x-4 y-10=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,445 |
一、(40 Points) As shown in Figure 1, in the acute triangle $\triangle ABC$, $\angle B > \angle C$, $M$ is the midpoint of side $BC$, $CD$ and $BE$ are the altitudes from $C$ and $B$ to sides $AB$ and $AC$ respectively, $K$ and $L$ are the midpoints of $ME$ and $MD$ respectively. If $KL$ intersects the line through point... | As shown in Figure 2, connect $D E$.
Since $B, D, E,$ and $C$ are concyclic, we have
$$
\begin{array}{l}
\angle A D E \\
=\angle B C A .
\end{array}
$$
In the right triangle $\triangle B D C$, by $D M=B M$, we get
$$
\begin{array}{l}
\angle B D M \\
=\angle D B M .
\end{array}
$$
Then $\angle A=180^{\circ}-\angle A B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,447 |
II. (40 points) Let positive real numbers $a, b, c$ satisfy $a+b+c=ab+bc+ca$.
If $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \geqslant k$ always holds, find the maximum value of $k$.
保留了源文本的换行和格式。 | II. The maximum value of $k$ is $\frac{4}{3}$.
Let $c=\frac{1}{2}$. Then $a+b+\frac{1}{2}=a b+\frac{1}{2} b+\frac{1}{2} a \Rightarrow a=\frac{b+1}{2 b-1}$. Let $b \rightarrow \frac{1}{2}\left(b>\frac{1}{2}\right)$. Then $a \rightarrow+\infty$. Hence $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \rightarrow \frac{4}{3}$.
T... | \frac{4}{3} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,448 |
For the integer $n(n \geqslant 2)$, does there exist positive integers $k_{1}, k_{2}, \cdots, k_{n}$ and primes $p_{1}, p_{2}, \cdots, p_{n}$, such that for any positive integer $m, k_{i} p_{i}^{m}+i(i=1,2, \cdots, n)$ contains at least one prime number? | Three, does not exist.
Take a sufficiently large positive integer $\iota$ such that $p_{i}^{t}>i(i=1,2$, $\cdots, n)$. Then
$$
k_{i} p_{i}^{m}+i=k_{i} p_{i}^{t}\left(p_{i}^{m-t}-1\right)+k_{i} p_{i}^{t}+i \text {. }
$$
By the choice of $t$, we can take a prime factor $s_{i}$ of $k_{i} p_{i}^{t}+i$ $\left(s_{i} \neq p_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,449 |
Four. (50 points) In a $2011 \times 2011$ grid, some cells are colored black such that no three black cells form an $\mathrm{L}$ shape (in any of the four possible orientations). What is the maximum number of cells that can be colored black? | First, color the first row, the third row, ..., and the last row. It is clearly satisfied, and the number of colored cells is $(n+1)(2 n+1)$.
Next, prove: If there are $(n+1)(2 n+1)+1$ cells colored, then there exists an L-shape.
When $n=1$, in a $3 \times 3$ grid, if 7 cells are colored, it is obvious that there exi... | (n+1)(2 n+1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,450 |
Example 1 Given $a b c \neq 0$, and $a+b+c=0$. Then the value of the algebraic expression $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | Solution 1 Original expression
$$
\begin{array}{l}
=\frac{-(b+c) a}{b c}+\frac{-(a+c) b}{a c}+\frac{-(a+b) c}{a b} \\
=-\left(\frac{a}{b}+\frac{a}{c}\right)-\left(\frac{b}{a}+\frac{b}{c}\right)-\left(\frac{c}{a}+\frac{c}{b}\right) \\
=3 .
\end{array}
$$
Solution 2 Given $a+b+c=0$, we can take
$$
a=b=1, c=-2 \text {. }... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,451 |
3. $\angle A 、 \angle B 、 \angle C$ are the interior angles of $\triangle A B C$.
Proposition 甲: $\angle A=\angle C$;
Proposition 乙: $\cos \frac{B}{2}=\sin C$.
Then 甲 is the $\qquad$ condition of 乙. | 3. Sufficient but not necessary.
When $\angle A=\angle C$, $\frac{\angle B}{2}+\angle C=\frac{\pi}{2}$.
Thus, $\cos \frac{B}{2}=\sin C$;
When $\angle A=\frac{3 \pi}{8}, \angle B=\frac{\pi}{12}, \angle C=\frac{13 \pi}{24}$, $\cos \frac{B}{2}=\sin C$, but $\angle A \neq \angle C$. | Sufficient but not necessary | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,455 |
4. Given $O$ as the origin, $M$ as a point on the moving chord $AB$ of the parabola $y^{2}=2 p x$ $(p>0)$. If $O A \perp O B, O M \perp A B$, denote the area of $\triangle O A B$ as $S$, and $O M=h$. Then the range of $\frac{S}{h}$ is . $\qquad$ | 4. $[2 p,+\infty)$.
Let $A\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right) 、 B\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right)$.
Since $O A \perp O B$, we have,
$$
\begin{array}{l}
y_{1} y_{2}=-4 p^{2}, \\
|A B| \geqslant\left|y_{1}-y_{2}\right|=\sqrt{y_{1}^{2}+y_{2}^{2}+8 p^{2}} \geqslant 4 p . \\
\text { Then } \frac{S}{h}=\frac{... | [2 p,+\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,456 |
5. The equation $x+y+z=2011$ satisfies $x<y<z$ to prove: the number of solutions $(x, y, z)$ is $\qquad$ groups. | 5. 336005 .
The equation $x+y+z=2011$ has $\mathrm{C}_{2010}^{2}$ sets of positive integer solutions. In each solution, $x, y, z$ cannot all be equal, and there are $3 \times 1005$ sets of solutions where two of $x, y, z$ are equal. Therefore, the number of solutions that meet the requirements is
$$
\frac{C_{2010}^{2}... | 336005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,457 |
6. For any real numbers $x, y, z$ not all zero, we have
$$
\begin{array}{l}
-6 x y + 18 z x + 36 y z . \\
\leqslant k\left(54 x^{2} + 41 y^{2} + 9 z^{2}\right) .
\end{array}
$$
Then the minimum value of the real number $k$ is | 6.1.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{c}
{\left[(x+y)^{2}+(z+x)^{2}+(y+z)^{2}\right]\left(4^{2}+8^{2}+10^{2}\right)} \\
\geqslant[4(x+y)+8(z+x)+10(y+z)]^{2} .
\end{array}
$$
Simplifying, we get $-11r$.
$$
54 x^{2}+41 y^{2}+9 z^{2} \geqslant-6 x y+18 z x+36 y z \text {. }
$$
When $x: y: z=1:... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,458 |
7. Given complex numbers $z_{1}, z_{2}, z_{3}$ satisfy
$$
\begin{array}{l}
\left|z_{1}\right| \leqslant 1,\left|z_{2}\right| \leqslant 1, \\
\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right| .
\end{array}
$$
Then the maximum value of $\left|z_{3}\right|$ is | 7. $\sqrt{2}$.
Notice that
$$
\begin{array}{l}
\left|2 z_{3}\right|-\left|z_{1}+z_{2}\right| \\
\leqslant\left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leqslant\left|z_{1}-z_{2}\right| .
\end{array}
$$
Then $2\left|z_{3}\right| \leqslant\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right|$
$$
\begin{array}{l}
\leqslant ... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,459 |
8. The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
a_{0}=0, a_{1}=1, a_{n}=2 a_{n-1}+a_{n-2}(n>1) \text {. }
$$
Then the number of $n$ that satisfies $2^{2011} \mid a_{n}$ is $\qquad$ . | 8. Infinitely many.
From the characteristic equation $x^{2}-2 x-1=0$, its two roots are $1+\sqrt{2}, 1-\sqrt{2}$.
From the initial conditions, the general term of the sequence is
$$
a_{n}=\frac{(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}}{2 \sqrt{2}} \text {. }
$$
By the binomial theorem, we have
$$
\begin{aligned}
& (1+\sqrt{... | Infinitely many | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,460 |
10. (20 points) For the curve $C_{1}$:
$$
3\left(x^{2}+2 y^{2}\right)^{2}=2\left(x^{2}+4 y^{2}\right)
$$
1 For every point $P$ except the origin, prove: there exists a line passing through $P$ that intersects the ellipse $x^{2}+2 y^{2}=2$ at two points $A$ and $B$, such that $\triangle A O P$ and $\triangle B O P$ are ... | 10. Let $P\left(x_{0}, y_{0}\right)\left(x_{0} y_{0} \neq 0\right), A\left(x_{1}, y_{1}\right)$, and $B\left(x_{2}, y_{2}\right)$, with the line $l_{A B}: y-y_{0}=k\left(x-x_{0}\right)$.
When combined with the equation $x^{2}+2 y^{2}=2$, we get
$$
\left(1+2 k^{2}\right) x^{2}+4 k\left(y_{0}-k x_{0}\right) x+2\left(y_{0... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,462 |
11. (20 points) Let $x, y, z \in(0,1)$, and $x^{2}+y^{2}+z^{2}=1$.
Determine the maximum value of $f=x+y+z-x y z$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 11. When $x=y=z=\frac{\sqrt{3}}{3}$, $f=\frac{8 \sqrt{3}}{9}$.
Next, we prove: $f \leqslant \frac{8 \sqrt{3}}{9}$.
Let $a=\sqrt{3} x, b=\sqrt{3} y, c=\sqrt{3} z$. Then $a^{2}+b^{2}+c^{2}=3$.
Hence, (1) $\Leftrightarrow 3(a+b+c)-a b c \leqslant 8$.
Notice that
$$
a b c=\frac{1}{3}\left[\sum a^{2}-\sum a\left(\sum a^{2... | \frac{8 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,463 |
Three, (50 points) Given $n\left(n \geqslant 3, n \in \mathbf{N}_{+}\right)$ pairwise coprime positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying: it is possible to appropriately add “ + ” or “ - ” to make the algebraic sum 0. Question: Does there exist a set of positive integers $b_{1}, b_{2}, \cdots, b_{n}$ (a... | When $n \geqslant 4$,
(1) If $b_{1}, b_{2}, \cdots, b_{n}$ contain two even numbers, then when
$k$ is even, $b_{1}+a_{1} k, b_{2}+a_{2} k, \cdots, b_{n}+a_{n} k$ will have two terms that are both even, and thus not coprime.
(2) If $b_{1}, b_{2}, \cdots, b_{n}$ contain at most one even number, then there are at least th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,467 |
Four. (50 points) Let $n \in \mathbf{N}_{+}, S=\{1,2, \cdots, 2 n\}$ have $k$ subsets $A_{1}, A_{2}, \cdots, A_{k}$ satisfying:
(1) For any $i \neq j(i, j=1,2, \cdots, k)$, $A_{i} \cap A_{j}$ has exactly an odd number of elements;
(2) For any $i(i=1,2, \cdots, k)$, $i \notin A_{i}$;
(3) If $i \in A_{j}$, then $j \in A_... | Four, $k_{\max }=2 n-1$.
First, the following $2 n-1$ sets satisfy conditions (1), (2), and (3):
$$
\begin{aligned}
A_{i} & =\{2 n-i, 2 n\}(i=1,2, \cdots, n-1, n+1, \cdots, 2 n-1), \\
A_{n} & =\{2 n\} .
\end{aligned}
$$
Next, we prove: $k \leqslant 2 n-1$.
If not, suppose the $2 n$ subsets $A_{1}, A_{2}, \cdots, A_{2 ... | 2n-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,468 |
Given real numbers $a, b, c, d$ satisfy $a^{4}+b^{4}=c^{4}+d^{4}=2011, a c+b d=0$. Find the value of $a b+c d$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Given $a c+b d=0 \Rightarrow a c=-b d$.
Raising both sides to the fourth power, we get $a^{4} c^{4}=b^{4} d^{4}$.
Since $c^{4}=2011-d^{4}, b^{4}=2011-a^{4}$, we have,
$$
\begin{array}{l}
a^{4}\left(2011-d^{4}\right)=\left(2011-a^{4}\right) d^{4} \\
\Rightarrow 2011 a^{4}-a^{4} d^{4}=2011 d^{4}-a^{4} d^{4} .
\end{array}... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,469 |
In trapezoid $A B C D$, $A B / / C D, A D$ $=49, B C=60, C D=B D=50$. Find the area of trapezoid $A B C D$. | Solve as shown in Figure 3, construct $D E \perp B C, B F \perp C D$, $A P \perp C D$, with the feet of the perpendiculars being $E, F, P$ respectively.
In the isosceles $\triangle B D C$, by the "three lines coincide" property, we have
$$
C E=\frac{1}{2} B C=30.
$$
In the right $\triangle D E C$, by the Pythagorean ... | 1536 \pm 24 \sqrt{97} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,470 |
If $n$ is a positive integer, then
$$
\begin{array}{l}
(\sec x \cdot \csc x)^{2 n}-\left(\sec ^{2 n} x+\csc ^{2 n} x\right) \\
\geqslant 2^{2 n}-2^{n+1} .
\end{array}
$$ | Proof Let $\tan x = t$. Then
$$
\begin{array}{l}
\left(\sec ^{2 n} x-1\right)\left(\csc ^{2 n} x-1\right) \\
=\left[\left(1+t^{2}\right)^{n}-1\right]\left[\left(1+\frac{1}{t^{2}}\right)^{n}-1\right] \\
=\left(n t^{2}+C_{n}^{2} t^{4}+\cdots+t^{2 n}\right)\left(n \frac{1}{t^{2}}+C_{n}^{2} \frac{1}{t^{4}}+\cdots+\frac{1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,471 |
310 four-digit number 2011 can be decomposed into the sum of squares of 14 positive integers, among which, 13 numbers form an arithmetic sequence. Write down this decomposition. | Given $1^{2}+3^{2}+\cdots+25^{2}>2011$, we know that the common difference of the arithmetic sequence is less than 2, i.e., the common difference is 1.
Let $2011=y^{2}+\sum_{k=-6}^{6}(x+k)^{2}$.
Then $2011=13 x^{2}+y^{2}+2 \sum_{k=1}^{6} k^{2}$.
Thus, $13 x^{2}+y^{2}=1829$.
Clearly, $x^{2}=\frac{1829-y^{2}}{13}$, so $x... | \begin{array}{l}
2011=16^{2}+\sum_{k=5}^{17} k^{2}, \\
2011=23^{2}+\sum_{k=4}^{16} k^{2}
\end{array} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,472 |
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